let
$$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\cdots\dfrac{1}{1}}}}}=\dfrac{m}{n}$$
where $m,n$ are positive integer numbers,and such $gcd(m,n)=1$.,and article $1998$ the fractional line on the left.
Find the value:
$m^2-mn-n^2$
and This problem is from this:http://www.mathchina.com/cgi-bin/topic.cgi?forum=5&topic=18459&show=0
and my try:
if Left hand have infinite the fractional line, we have
$$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\cdots}}}}=x$$ $$\Longrightarrow x=1+\dfrac{1}{x}\Longrightarrow x=\dfrac{1+\sqrt{5}}{2}$$
but left only have $1998$ the fractional line.so How find it ? Thank you
Let $x_0=1$ and $x_{k+1}=1+\frac1{x_k}$ for $n\in\Bbb N$. Suppose that $x_k=\frac{m_k}{n_k}$, where $m_k,n_k\in\Bbb Z^+$. Then
$$\frac{m_{k+1}}{n_{k+1}}=x_{k+1}=1+\frac1{x_k}=1+\frac{n_k}{m_k}=\frac{m_k+n_k}{m_k}\;.$$
Thus, we can get correct fractions by setting
$$\begin{align*} &m_0=n_0=1\;,\tag{1}\\ &m_{k+1}=m_k+n_k\;,\text{ and}\tag{2}\\ &n_{k+1}=m_k\;.\tag{3} \end{align*}$$
$(3)$ implies that $n_k=m_{k-1}$ for $k\ge 1$, so $(2)$ can be rewritten $$m_{k+1}=m_k+m_{k-1}\;,\tag{4}$$ with $m_0=1$ and $m_1=m_0+n_0=2$. Thus, $m_0=F_2$ and $m_1=F_3$, where $F_n$ is the $n$-th Fibonacci number. Now you can use known closed forms for the Fibonacci numbers to compute $m_{1998}=F_{2000}$ and $n_{1998}=m_{1997}=F_{1999}$.
Added: Then use the fact that
$$\begin{align*} F_{k+1}^2-F_kF_{k+1}-F_k^2&=(F_{k+1}-F_k)(F_{k+1}+F_k)-F_kF_{k+1}\\ &=F_{k-1}F_{k+2}-F_kF_{k+1}\\ &=F_{k-1}F_k+F_{k-1}F_{k+1}-F_kF_{k+1}\\ &=F_{k-1}F_{k+1}-F_k(F_{k+1}-F_{k-1})\\ &=F_{k-1}F_{k+1}-F_k^2\\ &=(-1)^k \end{align*}$$
by Cassini’s identity. (I’m assuming that the quadratic is supposed to be $n^2-mn-m^2$.)